Use of undefined constant STDIN - assumed 'STDIN'

Question

<?php
echo "Hello there. So I hear you're learning to be a PHP programmer!\n";
echo "Why don't you type in your name for me:\n";
$name = trim(fgets(STDIN));
echo "\nThanks, " . $name . ", it's really nice to meet you.\n\n";
?>

a am using php version 5.5.19 xampp

[What does the PHP error message “Notice: Use of undefined constant” mean?](http://stackoverflow.com/questions/2941169/what-does-the-php-error-message-notice-use-of-undefined-constant-mean) — Sean, May 16 '15 at 05:33
Are you running that in a web mode? [STDIN](http://php.net/manual/en/features.commandline.io-streams.php) is a CLI specific constant. — Ulver, May 16 '15 at 05:42
I think u should use define('STDIN',fopen("php://stdin","r")); in the at top of the php file. — user3419778, May 16 '15 at 06:17

score 3 · Answer 1 · answered May 16 '15 at 06:16

3

Assuming you are running from command line interface (CLI):

Define a STDIN constant at the top of your file:

define('STDIN', fopen('php://stdin', 'r'));

Or just replace STDIN constant with:

$name = trim(fgets(fopen('php://stdin', 'r')));

answered May 16 '15 at 06:16

Vahid Hallaji

6,676
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score 1 · Accepted Answer · edited Jun 11 '16 at 10:17

1

<?php

echo "Hello there. So I hear you're learning to be a PHP programmer!\n";

echo "Why don't you type in your name for me:\n";

$stdin = fopen('php://stdin', 'r');

$name=trim(fgets($stdin));

echo "\nThanks, " . $name . ", it's really nice to meet you.\n\n";

fclose($file);

?>

edited Jun 11 '16 at 10:17

Tamil Selvan C

19,232
12
49
68

answered May 16 '15 at 10:40

Tal

869
13
21

thanks this code work fine in cmd. – habibun May 17 '15 at 06:11

Use of undefined constant STDIN - assumed 'STDIN'

2 Answers2